What is the value of $\dfrac{d}{dx}\left(\dfrac{x^2-2x+3}{x+1}\right)$ at $x=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-2$ (Choice B) B $1$ (Choice C) C $-1$ (Choice D) D $-\dfrac12$
Solution: Let's first find the expression for $\dfrac{d}{dx}\left(\dfrac{x^2-2x+3}{x+1}\right)$ (i.e. for any input value $x$ ). Then, we can plug $x=1$ and evaluate. $\dfrac{x^2-2x+3}{x+1}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d d x ( x 2 − 2 x + 3 x + 1 ) = ( x + 1 ) d d x ( x 2 − 2 x + 3 ) − ( x 2 − 2 x + 3 ) d d x ( x + 1 ) ( x + 1 ) 2 The quotient rule = ( x + 1 ) ( 2 x − 2 ) − ( x 2 − 2 x + 3 ) ( 1 ) ( x + 1 ) 2 Differentiate ( x 2 − 2 x + 3 ) & ( x + 1 ) = 2 x 2 − 2 x + 2 x − 2 − x 2 + 2 x − 3 ( x + 1 ) 2 Expand = x 2 + 2 x − 5 ( x + 1 ) 2 \begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{x^2-2x+3}{x+1}\right) \\\\ &=\dfrac{(x+1)\dfrac{d}{dx}(x^2-2x+3)-(x^2-2x+3)\dfrac{d}{dx}(x+1)}{(x+1)^2} \gray{\text{The quotient rule}} \\\\ &=\dfrac{(x+1)(2x-2)-(x^2-2x+3)(1)}{(x+1)^2} \gray{\text{Differentiate }(x^2-2x+3)\text{ & }(x+1)} \\\\ &=\dfrac{2x^2-2x+2x-2-x^2+2x-3}{(x+1)^2}~~~ \gray{\text{Expand}} \\\\ &=\dfrac{x^2+2x-5}{(x+1)^2} \end{aligned} So we found that $\dfrac{d}{dx}\left(\dfrac{x^2-2x+3}{x+1}\right)=\dfrac{x^2+2x-5}{(x+1)^2}$. Now let's plug $x= 1$ : $\begin{aligned} &\phantom{=}\dfrac{( 1)^2+2( 1)-5}{(( 1)+1)^2} \\\\ &=\dfrac{1+2-5}{2^2} \\\\ &=-\dfrac12 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{x^2-2x+3}{x+1}\right)$ at $x=1$ is $-\dfrac12$.